0
$\begingroup$

If I have the matrix $A = \begin{pmatrix} 2 & 0 & 1 & -3 \\ 0 & 2 & 10 & 4 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}$ and I've calculated its Jordan normal form to be $J = \begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}$, how would I find the matrix $P$ such that $J = PAP^{-1}$?

$\endgroup$
  • $\begingroup$ The $P$ matrix is to be composed of vectors from Jordan chains corresponding to each of the Jordan blocks in the $J$ matrix. Did you compute some Jordan chains already for this problem (or do you have some worked examples to refer to?)? $\endgroup$ – Minus One-Twelfth Feb 14 at 14:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy